# A tricky equation of the first order

• Mar 7th 2011, 10:29 AM
A tricky equation of the first order
My teacher gave me a "bonus" assignment the other day, and I've been racking my brain trying to solve it. Here it is:

$\displaystyle xy'(ln(\frac{x}{y}) +1) = y(ln(\frac{x}{y}) - 1)$

So far I've figured that I should substitute $\displaystyle ln(\frac{x}{y})$ for $\displaystyle u(x)$, thus giving me $\displaystyle y = \frac{x}{e^{u(x)}}$, but from there I'm kind of lost. Any ideas?
• Mar 7th 2011, 10:35 AM
topsquark
Quote:

My teacher gave me a "bonus" assignment the other day, and I've been racking my brain trying to solve it. Here it is:

$\displaystyle xy'(ln(\frac{x}{y}) +1) = y(ln(\frac{x}{y}) - 1)$

So far I've figured that I should substitute $\displaystyle ln(\frac{x}{y})$ for $\displaystyle u(x)$, thus giving me $\displaystyle y = \frac{x}{e^{u(x)}}$, but from there I'm kind of lost. Any ideas?

I haven't done anything with it, but note that if you divide both sides by x you have an equation in y/x...

-Dan
• Mar 7th 2011, 10:41 AM
I'm sorry, my knowledge of differential equations in itself is quite shaky, but how would that help me? If I get what you're saying, is it that I get
$\displaystyle y'(ln(\frac{x}{y}) + 1) = \frac{x}{y}(ln(\frac{x}{y}) - 1)$?
If that's the case, are you suggesting I could substitute u(x) for $\displaystyle \frac{x}{y}$ instead?
• Mar 7th 2011, 10:53 AM
Ackbeet
The substitution $\displaystyle u = x/y$ does render the equation separable (the original DE is homogeneous). The problem is, the resulting integral in $\displaystyle u$ is horrendous. I suppose you could say, at that point, that you've "reduced the DE to quadratures", but if your professor is looking for a closed-form solution, that won't do.

The substitution $\displaystyle u=\ln(x/y)$ is much better. You have to translate the DE over to the $\displaystyle u$ domain:

$\displaystyle u'=\dfrac{y}{x}\cdot\dfrac{y-xy'}{y^{2}}=\dfrac{y-xy'}{xy}.$

Solve this for $\displaystyle y',$ and plug everything into the DE. A few things should cancel, leaving you with a much nicer separable equation.
• Mar 7th 2011, 11:00 AM
Ackbeet
Very cute problem, by the way! Thanks for posting!
• Mar 7th 2011, 11:05 AM
Thanks for the help. I'm trying it out right now, doing all the steps myself to see if I get it.
• Mar 7th 2011, 11:53 AM
topsquark
I just did the integral using Akbeet's method as well as using the homogeneous method and I'd rate them to be about the same level of difficulty.

For the "connoisseurs" out there, the homogeneous solution includes a nice little integral:
$\displaystyle \displaystyle \int \frac{ln(u)~du}{u}$ which I haven't seen done in a long time. (it's not particularly hard to do, just a nice little piece of work.)

-Dan
• Mar 7th 2011, 12:04 PM
OK, so I've gotten somewhere. Now I'm stuck on the actual separable differential equation.

I have

$\displaystyle \frac{2}{u + 1} = xu'$

$\displaystyle \frac{2}{u + 1} du = x dx$

but that approach seems to give me a completely different solution from what Wolfram Alpha gives me when I feed it the equation. So I figure I must be doing something wrong, since Wolfram Alpha's solution also seems more in tune with the final solution of the problem.
• Mar 7th 2011, 12:06 PM
Ackbeet
• Mar 7th 2011, 01:23 PM
OK, so I've hit one last roadblock and it's mighty frustrating. In the end I get

$\displaystyle \frac{u^2}{2} + u = 2ln(x) + c$

and Wolfram Alpha gives the solution

$\displaystyle \frac{1}{2}*ln^2(\frac{x}{y}) - ln(\frac{x}{y}) = 2ln(x) + c$

Which is basically the same thing with - instead of +. Should I show you step by step what I am doing? This is driving me nuts.
• Mar 7th 2011, 01:30 PM
Ackbeet
No, WolframAlpha is not giving you that solution, it's giving you this solution. Notice that the x and y in the WolframAlpha solution are flipped from what you have. Because the left-most logarithm is squared, the change in sign goes unnoticed. That, of course, doesn't happen with the second term.

I'm saying your solution is correct.
• Mar 7th 2011, 01:32 PM
Ackbeet
Incidentally, you could, if you wanted to, use the quadratic formula on log(x/y) and then solve for y. You get a multi-valued "function" then, but it is an explicit formula for y, which is nice.
• Mar 7th 2011, 01:39 PM